一起定义具有多个 API 版本的类 [英] Defining classes with several API versions together

问题描述

这显然不可能...

role Versioned {
    method version () {
        return self.^api;
    }
}

class WithApi:ver<0.0.1>:auth<github:JJ>:api<0> does Versioned {}
class WithApi:ver<0.0.1>:auth<github:JJ>:api<1> does Versioned {}

say WithApi:api<0>.new.version;
say WithApi:api<1>.new.version;

这与

==SORRY!=== Error while compiling /home/jmerelo/progs/perl6/my-perl6-examples/api-versioned.p6
Redeclaration of symbol 'WithApi'
at /home/jmerelo/progs/perl6/my-perl6-examples/api-versioned.p6:11
------> 1>:auth<github:JJ>:api<1> does Versioned⏏ {}

那么是否有可能在一个程序中使用具有不同apis、同名的类?

So is it even possible to use classes with different apis, same name in a single program?

更新:如果它们包含在不同的文件中，这是获得的错误:

Update: if they are included in different files, this is the error obtained:

P6M Merging GLOBAL symbols failed: duplicate definition of symbol WrongType

推荐答案

在这个例子中，有两件事造成了问题:

Two things are creating a problem in this example:

• class 默认是our，这会导致名称冲突
• 类的短名称在外部命名空间中相同，导致冲突

• class is by default our, which causes a name clash
• the short name of the class is the same in the outer namespace, causing a clash

如果我们稍微修改一下代码:

If we adapt the code slightly:

role Versioned {
    method version () {
        return self.^api;
    }
}

my constant one = my class WithApi:ver<0.0.1>:auth<github:JJ>:api<1> does Versioned {}
my constant two = my class WithApi:ver<0.0.1>:auth<github:JJ>:api<2> does Versioned {}

say one.version;  # 1
say two.version;  # 2

我确实发现 :api<0> 有一个错误.显然，这被认为等同于 no :api 设置，导致空字符串而不是 0.

I did find that there is a bug for :api<0>. Apparently this is considered to be equivalent to no :api setting, resulting in an empty string rather than 0.

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